Simplify; express your answer in exponential form. Assume $k\neq 0, t\neq 0$. $\dfrac{{(k^{4})^{-4}}}{{(kt)^{-5}}}$
Answer: To start, try working on the numerator and the denominator independently. In the numerator, we have ${k^{4}}$ to the exponent ${-4}$ . Now ${4 \times -4 = -16}$ , so ${(k^{4})^{-4} = k^{-16}}$ In the denominator, we can use the distributive property of exponents. ${(kt)^{-5} = (k)^{-5}(t)^{-5}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(k^{4})^{-4}}}{{(kt)^{-5}}} = \dfrac{{k^{-16}}}{{k^{-5}t^{-5}}}$ Break up the equation by variable and simplify. $\dfrac{{k^{-16}}}{{k^{-5}t^{-5}}} = \dfrac{{k^{-16}}}{{k^{-5}}} \cdot \dfrac{{1}}{{t^{-5}}} = k^{{-16} - {(-5)}} \cdot t^{- {(-5)}} = k^{-11}t^{5}$.